Question

下面的代码工作正常。我想处理错误情况，例如如果服务器的结果是null对象或搜索时出现一些数据库错误，我应该如何捕获错误并显示一些消息？

var nameIdMap = {};

        $('#selectAgent').typeahead({
        source: function (query, process) {
        var that = this;
            return $.ajax({
                dataType: "json",
                url: '/eBus/EbusinessAgentServlet',
                data: 'q=' + query +"&task=tab1&AgentId=1234",                                      
                type: 'POST',
                beforeSend: function() {
                //that.$element is a variable that stores the element the plugin was called on
                    that.$element.addClass('loading');
                },

                complete: function() {
                that.$element.removeClass('loading');
                },

                success: function (json) {
                console.log(json)
                    process(getOptionsFromJson(json));
                }
            });
        },
        minLength: 2,
        updater: function (item) {
            console.log('selected id'+nameIdMap[item]);
            return item;
        }
    });


});

 function getOptionsFromJson(json) {
$.each(json, function (i, v) {
    nameIdMap[v.fname + " "+ v.lname + " " + v.agentID] = v.agentID;
});

return $.map(json, function (n, i) {
    return n.fname +" " + n.lname+" " + n.agentID;
});
}

Answer 1

您可以使用fail()方法或statusCode选项进行错误处理，如下所示：

return $.ajax({
  dataType: "json",
  url: '/eBus/EbusinessAgentServlet',
  data: 'q=' + query +"&task=tab1&AgentId=1234",                                      
  type: 'POST',
  beforeSend: function() {
    //that.$element is a variable that stores the element the plugin was called on
    that.$element.addClass('loading');
  },
  complete: function() {
     that.$element.removeClass('loading');
  },
  success: function (json) {
     console.log(json)
     process(getOptionsFromJson(json));
  })
  .fail(function() {
    alert( "error" );
  });

return $.ajax({
  dataType: "json",
  url: '/eBus/EbusinessAgentServlet',
  data: 'q=' + query +"&task=tab1&AgentId=1234",                                      
  type: 'POST',
  beforeSend: function() {
    //that.$element is a variable that stores the element the plugin was called on
    that.$element.addClass('loading');
  },
  complete: function() {
     that.$element.removeClass('loading');
  },
  success: function (json) {
     console.log(json)
     process(getOptionsFromJson(json));
  }
  statusCode: {
    404: function() {
     alert( "page not found" );
  }
});

我希望它有所帮助。

bootstrap typeahead ajax错误处理

1 个答案: